Why does the reaction shift the left when adding [tex] H_{2}O [/tex]? 2. Relevant equations

3. The attempt at a solution

My first thought was that stress is being added to the right side of the equation and so equilibrium shifts left. This is incorrect because [tex] H_{2}O [/tex] is a liquid and therefore is not part of the equilibrium reaction (it has no concentration).

The addition of water does however dilute both sides of the equation. This is the explanation my teacher gave but going back over it, diluting the concentrations of the products and reactants would give you a larger [tex]K_{c}[/tex] which would shift it to the right.

At the molecular scale, the individual reactions don't stop just because the reaction is in equilibrium. They continue, but the rate going forward is the same as the rate going in reverse.

In other words, [itex]\mathrm{Co(H_2O)_6^{2+}}[/itex] ions get together and react with four chlorine ions each at the same rate that cobalt chloride ions get together and react with six water molecules.

If you had a lower [itex]\mathrm{Co(H_2O)_6^{2+}}[/itex] concentration, for example, the forward reaction would be less likely to occur because it would be harder to find a [itex]\mathrm{Co(H_2O)_6^{2+}}[/itex] ion to react.

This is incorrect because [tex] H_{2}O [/tex] is a liquid and therefore is not part of the equilibrium reaction (it has no concentration).

The addition of water does however dilute both sides of the equation. This is the explanation my teacher gave but going back over it,

are better than your first or third. Pedantically it is hardly correct to say water has no concentration, rather it has a constant concentration which essentially the rection does not change, therefore you can ignore water in considering the equilibrium. This is often expressed 'the concentration of water is conventionally set as 1M'